package com.dy.排序和搜索.中级.在排序数组中查找元素的第一个和最后一个位置;

/*
给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]
 */
public class Solution {
    //最蠢的暴力循环
    public int[] searchRange(int[] nums, int target) {
        int start = -1;
        int end = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                start = i;
                break;
            }
        }
        for (int i = 0; i < nums.length; i++) {

            if (nums[nums.length - 1 - i] == target) {
                end = nums.length - 1 - i;
                break;
            }
        }
        int[] res = {start, end};
        return res;

    }

    //二分法
    public int[] searchRange2(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int res[] = new int[2];
        res[0] = -1;
        res[1] = -1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                int i = mid, j = mid;
                while (i >= 0 && nums[i] == target) {
                    i--;
                }
                while (j < nums.length && nums[j] == target) {
                    j++;
                }
                res[0] = i+1;
                res[1] = j-1;
                break;
            }
            else if(nums[mid]<target){
                left = mid+1;
            }
            else if(nums[mid]>target){
                right = mid-1;
            }
        }
        return res;
    }
}
